Problem: Write the equation of a line that is perpendicular to $y=\dfrac{7}{5}x+6$ and that passes through the point $(2,-6)$.
Explanation: Getting started Key idea: The slopes of perpendicular lines are negative reciprocals of each other. Step 1: Find the slope Slope of the given line: ${\dfrac{7}{5}}$ So, the slope of the perpendicular line: $C{-\dfrac57}$ Step 2: Substitute the known point into linear equation The perpendicular line will have a slope of $C{-\dfrac{5}{7}}$ and pass through the point ${(2,-6)}$. Let's start from the point-slope form of the equation of the perpendicular line, then solve for $y$. [What is the point-slope form?] $\begin{aligned} y-({-6}) &= C{-\dfrac{5}{7}}(x-{2})\\\\\\ y+6 &= C{-\dfrac{5}{7}}x +\dfrac{10}{7} \\\\\\ y &= C{-\dfrac{5}{7}}x { -\dfrac{32}{7}} \end{aligned}$ Answer $y=C{-\dfrac{5}{7}}x {-\dfrac{32}{7}}$. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ $y$ $x$